Higher Maths Revision Notes Vectors Get Started

Vectors in three dimensions use scalar product to find the angle between two directed line segments know the terms: vector, magnitude (length), direction, scalar multiple, position vector, unit vector, directed line segment, component, scalar product determine the distance between two points in three dimensional space know and apply the equality fact know and apply the fact that if u and v are vectors that can be represented by parallel lines then u = kv where k is a constant and the converse know and apply the fact that if A, P and B are collinear points such that determine whether three points with given coordinates are collinear know and apply the basis vectors i, j, k know the scalar product facts: determine whether or not two vectors, in component form, are perpendicular a.b = |a| b| cos θ a.b = a 1 b 1 + a 2 b 2 + a 3 b 3 a.(b + c) = a.b + a.c

Test Yourself? B A O a b x y z u A vector u is represented by the directed line segment. The vector OA is called the position vector of A and is usually denoted by a. The vector OB is called the position vector of B and is usually denoted by b. From the diagram we see that a+ u = b. giving u = b – a i.e. = b – a.

Test Yourself? The converse is also true: If u and v are parallel then u = kv If AB = u and BC = v AND u = kv then, since they share a common point, B, AB and BC are line segments of the same line. i.e. A, B, and C are collinear. If vectors u and v are such that u = kv then u and v are parallel.

Test Yourself? Section Formula: If A, B and P are collinear with P between A and B, A B P km kn Using position vectors: n(p – a) = m(b – p)

Test Yourself? Vectors i, j, and k are unit vectors which run in the x, y and z-directions respectively. Any other vector can be expressed in terms of i, j, and k. The scalar (dot) product of two vectors a and b is defined as Since cos 90˚ = 0, when two vectors are perpendicular, their scalar product is zero.

Test Yourself? A B C θ Check BA and BC are tail-to-tail Find BA using BA = a – b Find BC using BC = c – b Find Find BA.BC using components. Calculate cos θ Hence calculate θ.

Test Yourself? Equality : When two vectors are perpendicular, their scalar product is zero. Perpendiculars :

reveal Three points on the ‘Cherry Picker’ have coordinates, A(4, –10, 8), B (1, 2, 4), and C(–3, –1, 16). Show that B is equidistant from A and C.

Three points on the ‘Cherry Picker’ have coordinates, A(4, –10, 8), B (1, 2, 4), and C(–3, –1, 16). Show that B is equidistant from A and C. AB = CB = 13. B is equidistant from A and C.

reveal The photographer was at P(1, 3, 5) The chimney of the castle was at C(2, 9, 20). A hill walker on the skyline was at H(6, 33, k). For what value of k is the photographer, chimney and hill walker in a straight line?

For P, C and H to be collinear, we require PC and PH to have a common point … P in this case, AND PC|| PH … so PH = aPC Now 5 = a ✕ 1 and 30 = a ✕ 6 … i.e. a = 5 So k – 5 = a ✕ 15 = 5 ✕ 15 = 75 So k = 80 when photographer, chimney and walker are in a straight line. The photographer was at P(1, 3, 5) The chimney of the castle was at C(2, 9, 20). A hill walker on the skyline was at H(6, 33, k). For what value of k is the photographer, chimney and hill walker in a straight line?

A P B A length of rigging starts at A(1, 2, 7) and runs to B(6, 12, 27). Find the point P that divides the length of rope, AB, in the ratio 3:2 reveal

A P B A length of rigging starts at A(1, 2, 7) and runs to B(6, 12, 27). Find the point P that divides the length of rope, AB, in the ratio 3:2 P is the point (4, 8, 19)

(a)u = 3j + 4k and v = 2i + 3j – 5k. Find u.v (b)ABC is an equilateral triangle. Find (c)ABCD is a square of side 1 unit. BD is a quarter circle centre A. EF cuts the square into two congruent rectangles. G lies on the arc BD. Calculate: A BC D E F G reveal

(a)u = 3j + 4k and v = 2i + 3j – 5k. Find u.v (b)PQRis an equilateral triangle of side 2 units. Find (c)ABCD is a square of side 1 unit. BD is a quarter circle centre A. EF cuts the square into two congruent rectangles. G lies on the arc BD. Calculate: A BC D E F G (a) u.v = (3j + 4k)(2i + 3j – 5k) = 6ji + 9jj – 15jk + 8ki + 12kj – 20kk = 9 – 20 = –11 (b) (c)

A(7, 4, 1) B(9, 6, 15) C(15, 3, 5) Calculate the size of angle ABC. reveal

A(7, 4, 1) B(9, 6, 15) C(15, 3, 5) Calculate the size of angle ABC.

Two vectors are defined as follows: (a)For what value of x is u = v? (b)For what value of x are u and v perpendicular? reveal

Two vectors are defined as follows: (a)For what value of x is u = v? (b)For what value of x are u and v perpendicular? (a) u = v when 2x = x + 2 and 4 – x = x This occurs when x = 2 (b) Vectors are perpendicular when u.v = 0